Titration of 10.0 mL of HCl(aq) with 0.100 mol/L...

Titration of 10.0 mL of HCl(aq) with 0.100 mol/L NaOH(aq) Trial 1 2 3 4 final burette reading (mL) 11.2 22.4 37.0 44.9 initial burette reading (mL) 0.0 11.2 22.4 33.6 volume of NaOH(aq) 11-2 11.2 added (mL) 11.2 1.? colour at endpoint grun green green grein 19. Calculate the average volume of titrant added. avl: 11.2 + 11.2 + 11.2 + 11.3 4 AV1: 11.2 mL

Transcript

00:01 Hello everyone in this question different trials has been given and we need to calculate first uh the moles of naoh added, moles of hcl added, molarity of hcl and the average molarity of hcl.

00:13 So first let's calculate the moles of naoh, moles of naoh for different trials we need to find it out.

00:22 So the first trial for trial one it would be equal to we know that how do we calculate the that is the volume into the concentration.

00:34 So moles is basically equal to volume into concentration this formula that we will use it over here.

00:44 So since it is a long question we'll directly put the values on this question on this formula.

00:49 So it would be equal to concentration that is equal to 0 .197 molar concentration into the volume that is equal to 0 .01710 liters.

01:00 So it comes out to be the number of moles that is 0 .0036 moles.

01:08 For trial two the concentration that is given to us is 0 .197 molar concentration into the volume that is equal to 0 .01741 liters.

01:24 So on calculating this we get the number of moles as 0 .00343 moles.

01:31 For trial three the concentration that is given to us is 0 .197 molar concentration into the volume that is 0 .01860 liters.

01:49 So in calculating we get the number of moles is equal to 0 .00331 moles.

01:54 So this was the first one that was the number of moles of naoh.

02:00 Now we need to calculate the moles of hcl.

02:06 So if we see at the balance chemical equation then the number of moles of hcl is equal to moles of naoh for all the trials that means for trial one, trial two and trial three the number of moles of hcl would be equal to number of moles of naoh for all three trials.

02:36 So now let's move to the next one that we supposed to calculate.

02:43 The next one is we need to calculate the molarity of hcl.

02:47 Molarity of hcl.

02:54 So the formula that we will use is that is moles of hcl divided by the volume of hcl...

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